Question: $-2rs + 7rt + 3r + 4 = 4s + 6$ Solve for $r$.
Answer: Combine constant terms on the right. $-2rs + 7rt + 3r + {4} = 4s + {6}$ $-2rs + 7rt + 3r = 4s + {2}$ Notice that all the terms on the left-hand side of the equation have $r$ in them. $-2{r}s + 7{r}t + 3{r} = 4s + 2$ Factor out the $r$ ${r} \cdot \left( -2s + 7t + 3 \right) = 4s + 2$ Isolate the $r$ $r \cdot \left( -{2s + 7t + 3} \right) = 4s + 2$ $r = \dfrac{ 4s + 2 }{ -{2s + 7t + 3} }$